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修改对sum的返回结果判断

hotfix/yujt_opt
zhangyuan 5 years ago
parent
commit
a209b8e563
  1. 4
      epdc-cloud-property/src/main/java/com/elink/esua/epdc/modules/project/service/impl/PropertyProjectServiceImpl.java

4
epdc-cloud-property/src/main/java/com/elink/esua/epdc/modules/project/service/impl/PropertyProjectServiceImpl.java

@ -209,8 +209,8 @@ public class PropertyProjectServiceImpl extends BaseServiceImpl<PropertyProjectD
@Override @Override
public ProjectDetailResultDTO getProjectDetailByIdV2(ProjectDetailFormDTO dto) { public ProjectDetailResultDTO getProjectDetailByIdV2(ProjectDetailFormDTO dto) {
ProjectDetailResultDTO projectDetailResultDTO = baseDao.selectDetailByIdV2(dto); ProjectDetailResultDTO projectDetailResultDTO = baseDao.selectDetailByIdV2(dto);
// 没有打分的情况需要额外判断 // 没有打分的情况需要额外判断,MySQL 5.0.3之前的版本,sum函数返回的是DOUBLE类型
if (projectDetailResultDTO != null && "0".equals(projectDetailResultDTO.getUserScore())) { if (projectDetailResultDTO != null && ("0.0".equals(projectDetailResultDTO.getUserScore()) || "0".equals(projectDetailResultDTO.getUserScore()))) {
ProjectDetailResultDTO score = baseDao.selectScoreByIdV2(dto); ProjectDetailResultDTO score = baseDao.selectScoreByIdV2(dto);
// 如果没有打分情况,赋空表示未打分 // 如果没有打分情况,赋空表示未打分
if (score == null) { if (score == null) {

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